Question: Let $f(x)=x^4+3x^3-x^2$. Find $f'(2)$. Choose 1 answer: Choose 1 answer: (Choice A) A $18$ (Choice B) B $36$ (Choice C) C $40$ (Choice D) D $64$
Let's first find the expression for $f'(x)$ and then evaluate it at $x=2$. According to the sum rule, the derivative of $x^4+3x^3-x^2$ is the sum of the derivatives of $x^4$, $3x^3$, and $-x^2$. The derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ For example, this is the derivative of the first term: $\dfrac{d}{dx}(x^4)=4x^3$ Here is the complete differentiation process: $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}(x^4+3x^3-x^2) \\\\ &=\dfrac{d}{dx}(x^4)+3\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(x^2)&&\gray{\text{Basic differentiation rules}} \\\\ &=4x^3+3\cdot3x^2-2x&&\gray{\text{The power rule}} \\\\ &=4x^3+9x^2-2x \end{aligned}$ So we found that $f'(x)=4x^3+9x^2-2x$. Plugging in $x=2$ and evaluating using the calculator, we find that $f'(2)=64$. In conclusion, $f'(2)=64$.